# Question #bf123

Aug 24, 2017

i) 2 loops
ii) 16 loops

#### Explanation:

The equation relating tension and speed of wave is given by:

$v = \sqrt{\frac{T}{\mu}}$
where: $T \text{ is the tension"; mu" is the linear density of the string}$

Since we know that $v = f \lambda$, we can rewrite the equation into:

$f \lambda = \sqrt{\frac{T}{\mu}}$

I)Tension increased by a factor of 4

Let $T$ be the initial tension of the string
$\therefore 4 T$ will be the final tension of the string

$4$loops are formed,
$L = 2 \lambda$

The string used is the same, thus the linear density, $\mu$ is a constant.
The frequency of the vibrator is also kept as the same, $f$ is another constant.

Rearranging
$f \lambda = \sqrt{\frac{T}{\mu}}$

${f}^{2} \mu = \frac{T}{\lambda} ^ 2$

since $f$ and $\mu$ are constants, their product is a constant too.

${T}_{i} / {\lambda}_{i}^{2} = {T}_{f} / {\lambda}_{f}^{2}$
subscript $i$ for initial, $f$ for final

$\frac{T}{\frac{L}{2}} ^ 2 = \frac{4 T}{\lambda} _ {f}^{2}$

$\frac{4 T}{4 \left({L}^{2} / 4\right)} = \frac{4 T}{\lambda} _ {f}^{2}$

$4 \left({L}^{2} / 4\right) = {\lambda}_{f}^{2}$

${\lambda}_{f}^{2} = {L}^{2}$

${\lambda}_{f} = L$

Therefore the number of loops formed will be 2.

II)Frequency increased by a factor of 4

Now, Tension instead of frequency becomes constant.

$f \lambda = \sqrt{\frac{T}{\mu}}$

Since Tension and Linear density remain the same, $\sqrt{\frac{T}{\mu}}$ is constant.

Thus,
${f}_{i} {\lambda}_{i} = {f}_{f} {\lambda}_{f}$

Let $f$ be the initial frequency,
$4 f$ will be the final frequency.

$4$loops are formed,
$L = 2 \lambda$
$\lambda = \frac{L}{2}$

$f \left(\frac{L}{2}\right) = \left(4 f\right) {\lambda}_{f}$

${\lambda}_{f} = f \frac{\frac{L}{2}}{4 f}$

${\lambda}_{f} = \frac{L}{8}$

Thefore, the number of loops is 16.